Question

If y = xⁿ⁻¹ log x then x² y₂ + (3 − 2n) xy₁ is equal to

विकल्प

• −(n − 1)² y

• (n − 1)²y

• −n²y

•  n²y

Answer 1

(a) −(n − 1)² y

Here,

\[y = x^{n - 1} \log x\]

\[ \Rightarrow y_1 = \left( n - 1 \right) x^{n - 2} \log x + \frac{x^{n - 1}}{x}\]

\[ \Rightarrow y_1 = \frac{\left( n - 1 \right) x^{n - 1} \log x + x^{n - 1}}{x}\]

\[ \Rightarrow x y_1 = \left( n - 1 \right)y + x^{n - 1} \]

\[ \Rightarrow x y_2 + y_1 = \left( n - 1 \right) y_1 + \left( n - 1 \right) x^{n - 2} \]

\[ \Rightarrow x y_2 + y_1 = \left( n - 1 \right) y_1 + \frac{\left( n - 1 \right) x^{n - 1}}{x}\]

\[ \Rightarrow x^2 y_2 + x y_1 = x\left( n - 1 \right) y_1 + \left( n - 1 \right) x^{n - 1} \]

\[ \Rightarrow x^2 y_2 + x y_1 = x\left( n - 1 \right) y_1 + \left( n - 1 \right)\left\{ x y_1 - \left( n - 1 \right)y \right\}\]

\[ \Rightarrow x^2 y_2 + x y_1 = x\left( n - 1 \right) y_1 + \left( n - 1 \right)x y_1 - \left( n - 1 \right)^2 y\]

\[ \Rightarrow x^2 y_2 + x y_1 = 2x\left( n - 1 \right) y_1 - \left( n - 1 \right)^2 y\]

\[ \Rightarrow x^2 y_2 + x y_1 - 2x\left( n - 1 \right) y_1 = - \left( n - 1 \right)^2 y\]

\[ \Rightarrow x^2 y_2 + x y_1 \left( 1 - 2n + 2 \right) = - \left( n - 1 \right)^2 y\]

\[ \Rightarrow x^2 y_2 + \left( 3 - 2n \right)x y_1 = - \left( n - 1 \right)^2 y\]